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    Could anyone help me with a differentiation question? Basically you see how the function (3x^2+5x-9)^1/3 can be differentiated to 1/3(6x+5)(3x^2+5x-9)^-2/3 so there is 2 brackets, could u help me differentiate ((x^3-4x+4)/2)^1/2 in the same way?

    Thank you for any help im really struggling.
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    Look this up on Wolfram alpha
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    (Original post by emperorCode)
    Look this up on Wolfram alpha
    I tried but they don't differentiate it into the way I want.
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    (Original post by mansnothot)
    Could anyone help me with a differentiation question? Basically you see how the function (3x^2+5x-9)^1/3 can be differentiated to 1/3(6x+5)(3x^2+5x-9)^-2/3 so there is 2 brackets, could u help me differentiate ((x^3-4x+4)/2)^1/2 in the same way?

    Thank you for any help im really struggling.
    Separate into 0.5^0.5 * (x^3-4x+4)^1/2, as 0.5^0.5 is a constant it shouldnt hinder ur ability to differentiate the other expression. Just remember to multiply by 0.5^0.5 after uve differentiated (x^3-4x+4)^1/2.
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    (Original post by Shaanv)
    Separate into 0.5^0.5 * (x^3-4x+4)^1/2, as 0.5^0.5 is a constant it shouldnt hinder ur ability to differentiate the other expression. Just remember to multiply by 0.5^0.5 after uve differentiated (x^3-4x+4)^1/2.
    So would this give me my differential function?
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    (Original post by mansnothot)
    So would this give me my differential function?
    This will give u the gradient function. Are u talking about a specific question or just in general in ur intial post?
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    (Original post by Shaanv)
    This will give u the gradient function. Are u talking about a specific question or just in general in ur intial post?
    no i just needed to differentiate that because I didnt know how to. If it's okay can I attempt to differentiate it and could you tell me if I'm correct?

    Thanks.
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    I went on this thinking it would be related to GCSE OCR Chemistry (Unit C3).

    My bad.
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    (Original post by RickiestRick)
    I went on this thinking it would be related to GCSE OCR Chemistry (Unit C3).

    My bad.
    SMH, change ur name to mortiest morty. 😂
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    (Original post by Shaanv)
    SMH, change ur name to mortiest morty. 😂
    But I'm arrogant, and that's exactly who Rick is
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    (Original post by RickiestRick)
    I went on this thinking it would be related to GCSE OCR Chemistry (Unit C3).

    My bad.
    That's such a dank unit!

    Wait you do new spec

    e . w
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    (Original post by mansnothot)
    Could anyone help me with a differentiation question? Basically you see how the function (3x^2+5x-9)^1/3 can be differentiated to 1/3(6x+5)(3x^2+5x-9)^-2/3 so there is 2 brackets, could u help me differentiate ((x^3-4x+4)/2)^1/2 in the same way?

    Thank you for any help im really struggling.
    nice username g
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    (Original post by perspirationting)
    nice username g
    Loooooooooool I'm done
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    (Original post by mansnothot)
    Could anyone help me with a differentiation question? Basically you see how the function (3x^2+5x-9)^1/3 can be differentiated to 1/3(6x+5)(3x^2+5x-9)^-2/3 so there is 2 brackets, could u help me differentiate ((x^3-4x+4)/2)^1/2 in the same way?

    Thank you for any help im really struggling.
    You use the chain rule for the first question and the quotient rule for the second.

    Not sure if this is allowed but I've attached an explanation of how to use the chain rule regarding the first question. Hope it makes sense and helps.

    Spoiler:
    Show


    Name:  IMG_20171026_224606692.jpg
Views: 17
Size:  493.0 KB

    For some bizarre reason, I can't upload the final answer, which is 1/3(6x + 5)(3x^2 + 5 - 9)^-2/3)

    Attached Images
      
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    Is it this...
    Spoiler:
    Show

    y = (½(x^3 - 4x + 4))^½
    = ½^½ * (x^3 - 4x + 4)^½
    dy/dx = ½^½ * ½(x^3 - 4x + 4)^-½ * (3x^2 - 4)
    = ½^3/2 * (x^3 - 4x + 4)^-½ * (3x^2 - 4)

    ∴ = ½^3/2 (x^3 - 4x + 4)^-½(3x^2 - 4)

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    (Original post by amkay)
    Is it this...
    Spoiler:
    Show



    y = (½(x^3 - 4x + 4))^½
    = ½^½ * (x^3 - 4x + 4)^½
    dy/dx = ½^½ * ½(x^3 - 4x + 4)^-½ * (3x^2 - 4)
    = ½^3/2 * (x^3 - 4x + 4)^-½ * (3x^2 - 4)

    ∴ = ½^3/2 (x^3 - 4x + 4)^-½(3x^2 - 4)



    I got up to this part ½^½ * ½(x^3 - 4x + 4)^-½ * (3x^2 - 4)
    but I don't understand why ½^½ * ½ becomes ½^3/2
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    (Original post by mansnothot)
    I got up to this part ½^½ * ½(x^3 - 4x + 4)^-½ * (3x^2 - 4)
    but I don't understand why ½^½ * ½ becomes ½^3/2
    I'm just adding powers so ½^½ * ½^1 = ½^(½ + 1) = ½^3/2
    just like how x^2 * x^3 = x^(2+3) = x^5
 
 
 
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